在删除至少包含一个“1”的子字符串“01”和“00”后，计算长度与给定字符串相同的二进制字符串

给定一个二进制字符串S，该任务是计数总二进制字符串由至少一个“1”的长度的后反复去除子串“10”的所有出现等于给定的字符串的长度，“00”，从给定的字符串.

例子：

Input: S = “111”
Output: 7
Explanation: Since there are no occurrences of “10” or “01” in the given string, length of the remaining string is 3. All possible binary strings of length 3 consisting of at least one set bit are {“001”, “010”, “011”, “100”, “101”, “110”, “111”}

Input: S = “0101”
Output: 3
Explanation: After deleting “10”, S = “01”. Therefore, length of remaining string is 2. Strings of length 2 consisting of at least one set bit are {“01”, “10”, “11”}

方法：这个想法是在从中删除所有形式为“10”和“00”的子字符串后计算给定字符串的长度。考虑剩下的；字符串的长度为N ，则由至少一个设置位组成的字符串总数将等于2 ^N -1 。

请按照以下步骤解决问题：

1. 初始化一个变量，比如count 。
2. 迭代字符串S的字符。对于每个字符，检查它是否为‘0’并且计数是否大于0 。如果发现为真，则将计数减1 。
3. 否则，如果当前字符为‘1’ ，则将count增加1 。
4. 完全遍历字符串，打印2 ^count – 1作为所需答案。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to count the strings
// consisting of at least 1 set bit
void countString(string S)
{
    // Initialize count
    long long count = 0;
 
    // Iterate through string
    for (auto it : S) {
 
        if (it == '0' and count > 0) {
            count--;
        }
        else {
            count++;
        }
    }
 
    // The answer is 2^N-1
    cout << ((1 << count) - 1) << "\n";
}
 
// Driver Code
int main()
{
 
    // Given string
    string S = "1001";
 
    // Function call
    countString(S);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
   
class GFG{
   
// Function to count the strings
// consisting of at least 1 set bit
static void countString(String S)
{
     
    // Initialize count
    int count = 0;
     
    // Iterate through string
    for(char it : S.toCharArray())
    {
        if (it == '0' && count > 0)
        {
            count--;
        }
        else
        {
            count++;
        }
    }
  
    // The answer is 2^N-1
    System.out.print((1 << count) - 1);
}
   
// Driver Code
public static void main(String[] args)
{
     
    // Given string
    String S = "1001";
  
    // Function call
    countString(S);
}
}
 
// This code is contributed by susmitakundugoaldanga

Python3

# Python3 program for the
# above approach 
 
# Function to count the
# strings consisting of
# at least 1 set bit
def countString(S):
   
    # Initialize count
    count = 0
     
    # Iterate through
    # string
    for i in S:
        if (i == '0' and
            count > 0):           
            count -= 1
         
        else:           
            count += 1
 
    # The answer is 2^N-1    
    print((1 << count) - 1) 
 
# Driver Code
if __name__ == "__main__":
 
    # Given string   
    S = "1001"
     
    # Function call   
    countString(S)
 
# This code is contributed by Virusbuddah_

C#

// C# program for the above approach
using System;
   
class GFG{
   
// Function to count the strings
// consisting of at least 1 set bit
static void countString(string S)
{
     
    // Initialize count
    int count = 0;
     
    // Iterate through string
    foreach(char it in S)
    {
        if (it == '0' && count > 0)
        {
            count--;
        }
        else
        {
            count++;
        }
    }
  
    // The answer is 2^N-1
    Console.Write((1 << count) - 1);
}
   
// Driver Code
public static void Main(string[] args)
{
     
    // Given string
    string S = "1001";
  
    // Function call
    countString(S);
}
}
 
// This code is contributed by chitranayal

Javascript

// javascript program for the above approach
 
    // Function to count the strings
    // consisting of at least 1 set bit
    function countString( S) {
 
        // Initialize count
        var count = 0;
 
        // Iterate through string
        for (var it =0;it< S.length; it++) {
            if (S[it] == '0' && count > 0) {
                count--;
            } else {
                count++;
            }
        }
 
        // The answer is 2^N-1
        document.write((1 << count) - 1);
    }
 
    // Driver Code
     
 
        // Given string
        var S = "1001";
 
        // Function call
        countString(S);
 
// This code contributed by umadevi9616

输出：

3

时间复杂度： O(L) 其中 L 是字符串的长度。
辅助空间： O(1)

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